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Ex 5.1, 10 - Find all points of discontinuity - Class 12 CBSE

Ex 5.1, 10 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.1, 10 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Ex 5.1, 10 - Chapter 5 Class 12 Continuity and Differentiability - Part 4

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Ex 5.1, 10 Find all points of discontinuity of f, where f is defined by 𝑓(π‘₯)={ β–ˆ(π‘₯+1, 𝑖𝑓 π‘₯β‰₯1@&π‘₯2+1 , 𝑖𝑓 π‘₯<1)─ Since we need to find continuity at of the function We check continuity for different values of x When x = 1 When x < 1 When x > 1 Case 1 : When x = 1 f(x) is continuous at π‘₯ =1 if L.H.L = R.H.L = 𝑓(1) if lim┬(xβ†’1^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’1^+ ) " " 𝑓(π‘₯)= 𝑓(1) Since there are two different functions on the left & right of 1, we take LHL & RHL . LHL at x β†’ 1 lim┬(xβ†’1^βˆ’ ) f(x) = lim┬(hβ†’0) f(1 βˆ’ h) = lim┬(hβ†’0) ((1βˆ’β„Ž)^2+1) = (1βˆ’0)^2+1 = 1^2+1 = 1 + 1 = 2 RHL at x β†’ 1 lim┬(xβ†’1^+ ) f(x) = lim┬(hβ†’0) f(1 + h) = lim┬(hβ†’0) (1+β„Ž)+1 = lim┬(hβ†’0) 2+β„Ž = 2 + 0 = 2 & 𝑓(1) = π‘₯+1 = 1+1 = 2 Hence, L.H.L = R.H.L = 𝑓(1) ∴ f is continuous at x = 1 Case 2 : When x < 1 For x < 1, f(x) = x2 + 1 Since this a polynomial It is continuous ∴ f(x) is continuous for x < 1 & 𝑓(1) = π‘₯+1 = 1+1 = 2 Hence, L.H.L = R.H.L = 𝑓(1) ∴ f is continuous at x = 1 Case 2 : When x < 1 For x < 1, f(x) = x2 + 1 Since this a polynomial It is continuous ∴ f(x) is continuous for x < 1 Case 3 : When x > 1 For x > 1, f(x) = x + 1 Since this a polynomial It is continuous ∴ f(x) is continuous for x > 1 Hence, there is no point of discontinuity Thus, f is continuous for all π’™βˆˆπ‘

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.